PHP show associative array content into JSON format - php

Question

i'm new in PHP language. I'm writing a PHP script to get information from my online database(created on my altervista personal webpage) and convert in JSON format.
The problem is about when i get elements from my database i can show them correctly using "echo" command, but when i parse associative array to "json_encode" function i can't see correctly elements.
Here is the code:
$sql = "SELECT * FROM People";
$result = $conn->query($sql);
$json_array = array();
if ($result->num_rows > 0) {
// output data of each row
$row = $result->fetch_assoc();
foreach($row as $key => $value) {
echo "" . $key . ": " . $value;
echo "<br>";
$json_array[] = array(''=>$key, ''=>$value);
}
} else {
echo "0 results";
}
echo json_encode($json_array);
$conn->close();
?>
The $result->fetch_assoc(); returns an associative array.
As i said before i can see correct information using echo commands.
The problem is when i use this:
$json_array[] = array(''=>$key, ''=>$value);
The output is as follow:
[{"":"Mark"},{"":"ABC"},{"":"25"}]
Basically it's showing me only the second parameter "=>$value", while the first parameter "=>$key" is ignored.
Can you tell me please what should i change in order to see also the first parameter in my output?
Thanks

Answer 1

The correct way would be to do that
$json_array[] = array( $key => $value );
instead of
$json_array[] = array(''=>$key, ''=>$value);
You are essentially adding an EMPTY string as key and the $key as value at first, and then you replace it with the $value as they share the EMPTY string key in the array.
Read more at: http://php.net/manual/en/language.types.array.php#language.types.array.syntax.array-func

Answer 2

First param is ignored because it has the same key as the second (the key is ''), you should write:
$json_array[] = array($key, $value); // Indexed array
Output will be:
[["key1", "value1"],["key2", "value2"],["key3", "value3"]]
OR
$json_array[] = array($key => $value); // Associative array
Output will be:
[{"key1": "value1"},{"key2": "value2"},{"key3": "value3"}]

Answer 3

Try:
sql = "SELECT * FROM People";
$result = $conn->query($sql);
$json_array = array();
if ($result->num_rows > 0) {
// output data of each row
while ($row = $result->fetch_assoc()) {
$json_array[] = $row;
}
} else {
echo "0 results";
}
echo json_encode($json_array);
$conn->close();
In your code you are parsing only the first db result